Question

when a satellite is a distance R from the center of Earth, the force due to gravity on the satellite is F. what is the force due to gravity on the satellite when its distance from the xenter of the earth is 3R?

Answer 1

Hello!

Do you know the equation that describes gravitational force between two masses?
(By the way... I haven't worked through it yet, but I think this will be a bit of algebra.)

Answer 2

\[F=G\frac{m_1 m_2}{r^2}\text{, where } G\approx 6.67\times 10^{-11}\]

Answer 3

We might as well write it out that the "F" there is at a distance of r. Oh! And I'll capitalize "R" so that it matches the question!
\[F_R=G\frac{m_1 m_2}{R^2}\text{, where } G\approx 6.67\times 10^{-11}\]

Answer 4

But in our next problem, our r isn't R anymore, it's (3R). So.......
\[G\frac{m_1 m_2}{(3R)^2}=?\]

Answer 5

If you were replying, I'd ask if you wanted to name the gravitational force for 3R or not. But since you're not here, I'll name it \[F_{3R}\]

Answer 6

\[G\frac{m_1 m_2}{(3R)^2}=F_{3R}\]Now we have two distinquished forces we can keep track of:\[F_R\text{ and }F_{3R}\]

Answer 7

distinct* - not distinguished, and not distinquished either! My bad.

Answer 8

So what's the difference? Again, another choice I would give you - algebraic vs. logic, to solve the problem.

Answer 9

Logic first.
What changed from \[F_R \text{ to }F_{3R}?\]The R turned to 3R. It was squared, and in the denominator. Since the 3 was squared and in the denominator, multiplying the term, we can say it's like\[\frac{1}{3^2}=\frac{1}{9}\]multiplying the term.
So\[F_R\frac{1}{9}=F_{3R}\text{ since we multiplied } F_R\text{ by }\frac{1}{9}\text{ to get }F_{3R}\]

Answer 10

Algebraically, well, I don't know really. I guess that's the only way I can think of right now. But it does what you need it to! If you know \[F_R\]you can very easily get\[F_{3R}\]